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Negative and positive

  1. Jun 2, 2008 #1
    given acceleration [tex]a = 1 + ln x[/tex]
    i can find that [tex]\Delta v^2 = 2xlnx[/tex] and since it is given that when [tex]t = 0, x = 1,v = 0[/tex]
    [tex]\therefore v^2 = 2xlnx[/tex]
    however i have been asked to prove [tex]v > 0 \; when \;t > 0[/tex] and i have no idea how to explain it in mathematcial terms, can anyone please give any suggestion?
  2. jcsd
  3. Jun 2, 2008 #2

    Gib Z

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    [tex]a = \frac{dv}{dt} > 0[/tex] since 1+ ln x is greater than ...
  4. Jun 2, 2008 #3
    I think your math is wrong calc_jy. Invoking work-energy theorem

    [tex]v^2(x) - v^2(1) = 2\int_{1}^{x} a(s)ds \Rightarrow [/tex]

    [tex]v^2(x) - 0 = 2\int_{1}^{x} (1+\ln s)ds \Rightarrow [/tex]

    [tex]v^2 = 2(s + 1/s) |_{1}^{x} \Rightarrow [/tex]

    [tex]v^2 = 2(x+1/x - 1 - 1) \Rightarrow[/tex]

    [tex]v^2 = 2(x + 1/x - 2)[/tex]

    But anyway you don't want that expression, just do what Gib said.
  5. Jun 2, 2008 #4
    The integral of ln(x) is not 1/x. You have that backwards.
  6. Jun 3, 2008 #5
    why is [tex]a\geq0 [/tex] as lnx can range from -infinite to infinite and what working do i need to actually prove that v>0 as t>0
    Last edited: Jun 3, 2008
  7. Jun 3, 2008 #6
    It seems to me that you want to solve this DE

    I'm assuming that x is a function of t.


    For the life of me, I can't remember how to solve that. I'll look it up later.
  8. Jun 3, 2008 #7

    Gib Z

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    I'm not too sure about that working, but I do know for sure v^2 = 2x log (x) . We can see it after seeing a = d/d(x) [ v^2/2] , and then integrating both sides.

    Well, yes the function log x alone does have that range, but remember: v^2 = 2x log x. The quantity on the left side is positive. The quantity on the right hand side must also be positive. That means x must be greater than or equal to 1. Which means log x must be greater than zero, which means a = 1 + log x must also always be greater than 1.

    a = dv/dt.

    dv/dt is strictly positive. Also, t is a strictly positive quantity. Hence v is also > 0.
  9. Jun 3, 2008 #8
    My bad! That's a terrible mistake to make. Well in the bizarro world were all derivatives are antiderivatives and all functions are exponentials, I'd be fine... but as we live in the normal world my blunder was inexcusable.:blushing:
  10. Jun 4, 2008 #9
    thanks !
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